[VISUAL] Based on the structures shown above, which of the following statements identifies the compo — Bonding Chemistry Question

This question continues our exploration of intermolecular forces, this time comparing constitutional isomers to see how structural differences alter macroscopic physical properties like boiling points.

STEPS:

1. Analyze the molecular formulas of both compounds:
* Both Compound 1 (N,N-dimethylethanamine) and Compound 2 (butan^-1-amine) are constitutional isomers with the same molecular formula ($\text{C}_4\text{H}_{11}\text{N}$) and an identical molar mass of approximately $73\text{ g/mol}$.
* Because they have the same size, shape, and number of electrons, their London dispersion forces (polarizabilities) are extremely similar.
2. Identify the structural differences in functional groups:
* Compound 1 is a tertiary amine. The central nitrogen atom is covalently bonded to three carbon atoms (one ethyl group and two methyl groups). Crucially, there are no hydrogen atoms bonded directly to the nitrogen atom (no $\text{N–H}$ bonds are present).
* Compound 2 is a primary amine. The nitrogen atom is covalently bonded to a butyl carbon chain and two hydrogen atoms (possessing two highly polar $\text{N–H}$ bonds).
3. Evaluate the capacity for hydrogen bonding:
* A hydrogen bond is a strong intermolecular attraction that occurs when a hydrogen atom covalently bonded to a highly electronegative atom ($\text{F}$, $\text{O}$, or $\text{N}$) is attracted to the lone pair of an electronegative atom on a neighboring molecule.
* Because Compound 2 contains polar $\text{N–H}$ bonds, its molecules can form a network of hydrogen bonds with each other in the liquid state.
* Because Compound 1 lacks any hydrogen atoms bonded to its nitrogen, its molecules cannot form hydrogen bonds with one another. They can only interact via weaker dipole-dipole attractions and London dispersion forces.
4. Relate intermolecular force (IMF) strength to boiling point:
* Boiling requires supplying enough thermal energy to overcome the intermolecular attractions holding molecules together in the liquid phase.
* Since hydrogen bonds are significantly stronger than ordinary dipole-dipole attractions, the molecules of Compound 2 are held together much more tightly than those of Compound 1.
5. Conclude:
* Consequently, Compound 2 requires more energy to boil and has the higher boiling point, which identifies Option D as the correct answer.

*

WHY_OTHERS_WRONG:

• Option A is incorrect: This option incorrectly claims that Compound 1 has a higher boiling point. While Compound 1 does have dipole-dipole forces, they are much weaker than the hydrogen bonding interactions experienced by Compound 2.
• Option B is incorrect: This option incorrectly states that Compound 1 is the one that forms hydrogen bonds. As shown in its structural formula, Compound 1 has no $\text{N–H}$ bonds, making it impossible for its molecules to form hydrogen bonds with one another.
• Option C is incorrect: This option correctly identifies Compound 2 as having the higher boiling point, but provides a faulty explanation. Because Compound 1 and Compound 2 are structural isomers with the same molecular formula, they have nearly identical polarizabilities and London dispersion forces. The difference in their boiling points is driven by hydrogen bonding, not a difference in dispersion forces.